A large, thermally isolated container is initially partitioned into two volumes,

Question

A large, thermally isolated container is initially partitioned into two volumes, V0 and 2V0. Gaseous helium and neon, which you can treat as ideal gases, have pressures P0 and 3P0, respectively, in the two regions. The temperature is uniform throughout the entire volume and has initial value T0. A thin aluminum partition between the chambers is allowed to slide and equilibrium is established. Answer the following in terms of V0, P0, and T0.

a. What is the final temperature?

b. What is the final pressure?

c. What is the total change in the entropy?

Explanation / Answer

as the container is thermally isolated, there is no exchange of heat with outside.

also as there is no change in volume, work done=zero.

hence total change in internal energy=0

==>there is no change in temeprature of the system

hence final temperature=initial temperature=T0

part b:

for the first partition, number of moles=P0*V0/(R*T0)

for second partition, number of moles=3*P0*2*V0/(R*T0)=6*P0*V0/(R*T0)

when the aluminium partion is lifted,

total volume=V0+2*V0=3*V0

temperature=T0

total number of moles=7*P0*V0/(R*T0)

then using the ideal gas law,

final pressure=(7*P0*V0/(R*T0))*R*T0/(3*V0)=(7/3)*P0

hence final pressure is (7/3)*P0.

part c:

as in case of joule expansion, change in entropy is n*R*ln(2) where n=number of moles,

here total change in entropy=(7*P0*V0/R*T0)*R*ln(2).

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